4 Answers
4

You are correct that your first vector needs to have dot product zero with $(4,7,-9)$. Just pick any $x,y$ you like and solve for $z$. I will pick $x=9,y=0$ and find $z=4$ works, so we have $(9,0,4)$ Now you can either do the same with dot products with both vectors, or you can take the cross product, which is guaranteed to be perpendicular to both. So take $(4,7,-9) \times (9,0,4)$ getting $(28, -97, -63)$. I admit, I used Alpha do do the work.

@Dimitri Topaloglou: If you don't need the two vectors to be perpendicular to each other, you can repeat my construction of the first vector to get, for example, $(4,-7,0)$ or $(0,7,9)$ more easily.
–
Ross MillikanMar 19 '13 at 23:21

Take vector $n = (x,y,z)$. As you said by yourself, dot product should vanish. So
$$
(4,7,-9) \cdot (x,y,z) = 4x+7y-9z = 0
$$
As you might see, all points that lie on the plane $4x+7y-9z = 0$ satisfy the condition of perpendicularity. If you want two linear independent vectors, just pick two different points. So, pick any two different triples $n_1 = (x_1, y_1, (4x_1+7y_1)/9)$ and $n_2 = (x_2, y_2, (4x_2+7y_2)/9)$ where $(x_1, y_1) \ne (x_2, y_2)$, and you're done.